## My Little 2021 Mathematics A-to-Z: Analysis

I’m fortunate this week to have another topic suggested again by Mr Wu, blogger and Singaporean mathematics tutor. It’s a big field, so forgive me not explaining the entire subject.

# Analysis.

Analysis is about proving why the rest of mathematics works. It’s a hard field. My experience, a typical one, included crashing against real analysis as an undergraduate and again as a graduate student. It turns out mathematics works by throwing a lot of $\epsilon$ symbols around.

Let me give an example. If you read pop mathematics blogs you know about the number represented by $0.999999\cdots$. You’ve seen proofs, some of them even convincing, that this number equals 1. Not a tiny bit less than 1, but exactly 1. Here’s a real-analysis treatment. And — I may regret this — I recommend you don’t read it. Not closely, at least. Instead, look at its shape. Look at the words and symbols as graphic design elements, and trust that what I say is not nonsense. Resume reading after the horizontal rule.

It’s convenient to have a name for the number $0.999999\cdots$. I’ll call that $r$, for “repeating”. 1 we’ll call 1. I think you’ll grant that whatever r is, it can’t be more than 1. I hope you’ll accept that if the difference between 1 and r is zero, then r equals 1. So what is the difference between 1 and r?

Give me some number $\epsilon$. It has to be a positive number. The implication in the letter $\epsilon$ is that it’s a small number. This isn’t actually required in general. We expect it. We feel surprise and offense if it’s ever not the case.

I can show that the difference between 1 and r is less than $\epsilon$. I know there is some smallest counting number N so that $\epsilon > \frac{1}{10^{N}}$. For example, say $\epsilon$ is 0.125. Then we can let N = 1, and $0.125 > \frac{1}{10^{1}}$. Or suppose $\epsilon$ is 0.00625. But then if N = 3, $0.00625 > \frac{1}{10^{3}}$. (If $\epsilon$ is bigger than 1, let N = 1.) Now we have to ask why I want this N.

Whatever the value of r is, I know that it is more than 0.9. And that it is more than 0.99. And that it is more than 0.999. In fact, it’s more than the number you get by truncating r after any whole number N of digits. Let me call $r_N$ the number you get by truncating r after N digits. So, $r_1 = 0.9$ and $r_2 = 0.99$ and $r_5 = 0.99999$ and so on.

Since $r > r_N$, it has to be true that $1 - r < 1 - r_N$. And since we know what $r_N$ is, we can say exactly what $1 - r_N$ is. It's $\frac{1}{10^{N}}$. And we picked N so that $\frac{1}{10^{N}} < \epsilon$. So $1 - r < 1 - r_N = \frac{1}{10^{N}} < \epsilon$. But all we know of $\epsilon$ is that it's a positive number. It can be any positive number. So $1 - r$ has to be smaller than each and every positive number. The biggest number that’s smaller than every positive number is zero. So the difference between 1 and r must be zero and so they must be equal.

That is a compelling argument. Granted, it compels much the way your older brother kneeling on your chest and pressing your head into the ground compels. But this argument gives the flavor of what much of analysis is like.

For one, it is fussy, leaning to technical. You see why the subject has the reputation of driving off all but the most intent mathematics majors. If you get comfortable with this sort of argument it’s hard to notice anymore.

For another, the argument shows that the difference between two things is less than every positive number. Therefore the difference is zero and so the things are equal. This is one of mathematics’ most important tricks. And another point, there’s a lot of talk about $\epsilon$. And about finding differences that are, it usually turns out, smaller than some $\epsilon$. (As an undergraduate I found something wasteful in how the differences were so often so much less than $\epsilon$. We can’t exhaust the small numbers, though. It still feels uneconomic.)

Something this misses is another trick, though. That’s adding zero. I couldn’t think of a good way to use that here. What we often get is the need to show that, say, function $f$ and function $g$ are equal. That is, that they are less than $\epsilon$ apart. What we can often do is show that $f$ is close to some related function, which let me call $f_n$.

I know what you’re suspecting: $f_n$ must be a polynomial. Good thought! Although in my experience, it’s actually more likely to be a piecewise constant function. That is, it’s some number, eg, “2”, for part of the domain, and then “2.5” in some other region, with no transition between them. Some other values, even values not starting with “2”, in other parts of the domain. Usually this is easier to prove stuff about than even polynomials are.

But get back to $g_n$. It’s got the same deal as $f_n$, some approximation easier to prove stuff about. Then we want to show that $g$ is close to some $g_n$. And then show that $f_n$ is close to $g_n$. So — watch this trick. Or, again, watch the shape of this trick. Read again after the horizontal rule.

The difference $| f - g |$ is equal to $| f - f_n + f_n - g |$ since adding zero, that is, adding the number $( -f_n + f_n )$, can’t change a quantity. And $| f - f_n + f_n - g |$ is equal to $| f - f_n + f_n -g_n + g_n - g |$. Same reason: $( -g_n + g_n )$ is zero. So:

$| f - g | = |f - f_n + f_n -g_n + g_n - g |$

Now we use the “triangle inequality”. If a, b, and c are the lengths of a triangle’s sides, the sum of any two of those numbers is larger than the third. And that tells us:

$|f - f_n + f_n -g_n + g_n - g | \le |f - f_n| + |f_n - g_n| + | g_n - g |$

And then if you can show that $| f - f_n |$ is less than $\frac{1}{3}\epsilon$? And that $| f_n - g_n |$ is also $\frac{1}{3}\epsilon$? And you see where this is going for $| g_n - g |$? Then you’ve shown that $| f - g | \le \epsilon$. With luck, each of these little pieces is something you can prove.

Don’t worry about what all this means. It’s meant to give a flavor of what you do in an analysis course. It looks hard, but most of that is because it’s a different sort of work than you’d done before. If you hadn’t seen the adding-zero and triangle-inequality tricks? I don’t know how long you’d need to imagine them.

There are other tricks too. An old reliable one is showing that one thing is bounded by the other. That is, that $f \le g$. You use this trick all the time because if you can also show that $g \le f$, then those two have to be equal.

The good thing — and there is good — is that once you get the hang of these tricks analysis starts to come together. And even get easier. The first course you take as a mathematics major is real analysis, all about functions of real numbers. The next course in this track is complex analysis, about functions of complex-valued numbers. And it is easy. Compared to what comes before, yes. But also on its own. Every theorem in complex analysis named after Augustin-Louis Cauchy. They all show that the integral of your function, calculated along a closed loop, is zero. I exaggerate by $\epsilon$.

In grad school, if you make it, you get to functional analysis, which examines functions on functions and other abstractions like that. This, too, is easy, possibly because all the basic approaches you’ve seen several courses over. Or it feels easy after all that mucking around with the real numbers.

This is not the entirety of explaining how mathematics works. Since all these proofs depend on how numbers work, we need to show how numbers work. How logic works. But those are subjects we can leave for grad school, for someone who’s survived this gauntlet.

I hope to return in a week with a fresh A-to-Z essay. This week’s essay, and all the essays for the Little Mathematics A-to-Z, should be at this link. And all this year’s essays, and all A-to-Z essays from past years, should be at this link. Thank you once more for reading.

## My 2018 Mathematics A To Z: Limit

I got an irresistible topic for today’s essay. It’s courtesy Peter Mander, author of Carnot Cycle, “the classical blog about thermodynamics”. It’s bimonthly and it’s one worth waiting for. Some of the essays are historical; some are statistical-mechanics; many are mixtures of them. You could make a fair argument that thermodynamics is the most important field of physics. It’s certainly one that hasn’t gotten the popularization treatment it deserves, for its importance. Mander is doing something to correct that.

# Limit.

It is hard to think of limits without thinking of motion. The language even professional mathematicians use suggests it. We speak of the limit of a function “as x goes to a”, or “as x goes to infinity”. Maybe “as x goes to zero”. But a function is a fixed thing, a relationship between stuff in a domain and stuff in a range. It can’t change any more than January, AD 1988 can change. And ‘x’ here is a dummy variable, part of the scaffolding to let us find what we want to know. I suppose ‘x’ can change, but if we ever see it, something’s gone very wrong. But we want to use it to learn something about a function for a point like ‘a’ or ‘infinity’ or ‘zero’.

The language of motion helps us learn, to a point. We can do little experiments: if $f(x) = \frac{sin(x)}{x}$, then, what should we expect it to be for x near zero? It’s irresistible to try out the calculator. Let x be 0.1. 0.01. 0.001. 0.0001. The numbers say this f(x) gets closer and closer to 1. That’s good, right? We know we can’t just put in an x of zero, because there’s some trouble that makes. But we can imagine creeping up on the zero we really wanted. We might spot some obvious prospects for mischief: what if x is negative? We should try -0.1, -0.01, -0.001 and so on. And maybe we won’t get exactly the right answer. But if all we care about is the first (say) three digits and we try out a bunch of x’s and the corresponding f(x)’s agree to those three digits, that’s good enough, right?

This is good for giving an idea of what to expect a limit to look like. It should be, well, what it really really really looks like a function should be. It takes some thinking to see where it might go wrong. It might go to different numbers based on which side you approach from. But that seems like something you can rationalize. Indeed, we do; we can speak of functions having different limits based on what direction you approach from. Sometimes that’s the best one can say about them.

But it can get worse. It’s possible to make functions that do crazy weird things. Some of these look like you’re just trying to be difficult. Like, set f(x) equal to 1 if x is rational and 0 if x is irrational. If you don’t expect that to be weird you’re not paying attention. Can’t blame someone for deciding that falls outside the realm of stuff you should be able to find limits for. And who would make, say, an f(x) that was 1 if x was 0.1 raised to some power, but 2 if x was 0.2 raised to some power, and 3 otherwise? Besides someone trying to prove a point?

Fine. But you can make a function that looks innocent and yet acts weird if the domain is two-dimensional. Or more. It makes sense to say that the functions I wrote in the above paragraph should be ruled out of consideration. But the limit of $f(x, y) = \frac{x^3 y}{x^6 + y^2}$ at the origin? You get different results approaching in different directions. And the function doesn’t give obvious signs of imminent danger here.

We need a better idea. And we even have one. This took centuries of mathematical wrangling and arguments about what should and shouldn’t be allowed. This should inspire sympathy with Intro Calc students who don’t understand all this by the end of week three. But here’s what we have.

I need a supplementary idea first. That is the neighborhood. A point has a neighborhood if there’s some open set that contains it. We represent this by drawing a little blob around the point we care about. If we’re looking at the neighborhood of a real number, then this is a little interval, that’s all. When we actually get around to calculating, we make these neighborhoods little circles. Maybe balls. But when we’re doing proofs about how limits work, or how we use them to prove things, we make blobs. This “neighborhood” idea looks simple, but we need it, so here we go.

So start with a function, named ‘f’. It has a domain, which I’ll call ‘D’. And a range, which I want to call ‘R’, but I don’t think I need the shorthand. Now pick some point ‘a’. This is the point at which we want to evaluate the limit. This seems like it ought to be called the “limit point” and it’s not. I’m sorry. Mathematicians use “limit point” to talk about something else. And, unfortunately, it makes so much sense in that context that we aren’t going to change away from that.

‘a’ might be in the domain ‘D’. It might not. It might be on the border of ‘D’. All that’s important is that there be a neighborhood inside ‘D’ that contains ‘a’.

I don’t know what f(a) is. There might not even be an f(a), if a is on the boundary of the domain ‘D’. But I do know that everything inside the neighborhood of ‘a’, apart from ‘a’, is in the domain. So we can look at the values of f(x) for all the x’s in this neighborhood. This will create a set, in the range, that’s known as the image of the neighborhood. It might be a continuous chunk in the range. It might be a couple of chunks. It might be a single point. It might be some crazy-quilt set. Depends on ‘f’. And the neighborhood. No matter.

Now I need you to imagine the reverse. Pick a point in the range. And then draw a neighborhood around it. Then pick out what we call the pre-image of it. That’s all the points in the domain that get matched to values inside that neighborhood. Don’t worry about trying to do it; that’s for the homework practice. Would you agree with me that you can imagine it?

I hope so because I’m about to describe the part where Intro Calc students think hard about whether they need this class after all.

All right. Then I want something in the range. I’m going to call it ‘L’. And it’s special. It’s the limit of ‘f’ at ‘a’ if this following bit is true:

Think of every neighborhood you could pick of ‘L’. Can be big, can be small. Just has to be a neighborhood of ‘L’. Now think of the pre-image of that neighborhood. Is there always a neighborhood of ‘a’ inside that pre-image? It’s okay if it’s a tiny neighborhood. Just has to be an open neighborhood. It doesn’t have to contain ‘a’. You can allow a pinpoint hole there.

If you can always do this, however tiny the neighborhood of ‘L’ is, then the limit of ‘f’ at ‘a’ is ‘L’. If you can’t always do this — if there’s even a single exception — then there is no limit of ‘f’ at ‘a’.

I know. I felt like that the first couple times through the subject too. The definition feels backward. Worse, it feels like it begs the question. We suppose there’s an ‘L’ and then test these properties about it and then if it works we say we’re done? I know. It’s a pain when you start calculating this with specific formulas and all that, too. But supposing there is an answer and then learning properties about it, including whether it can exist? That’s a slick trick. We can use it.

Thing is, the pain is worth it. We can calculate with it and not have to out-think tricky functions. It works for domains with as many dimensions as you need. It works for limits that aren’t inside the domain. It works with domains and ranges that aren’t real numbers. It works for functions with weird and complicated domains. We can adapt it if we want to consider limits that are constrained in some way. It won’t be fooled by tricks like I put up above, the f(x) with different rules for the rational and irrational numbers.

So mathematicians shrug, and do enough problems that they get the hang of it, and use this definition. It’s worth it, once you get there.

This and other Fall 2018 Mathematics A-To-Z posts can be read at this link. And I’m still taking nominations for discussion topics, if you’d like to see mathematics terms explained. I know I would.

## Wronski’s Formula For Pi: My Boring Mistake

Previously:

So, I must confess failure. Not about deciphering Józef Maria Hoëne-Wronski’s attempted definition of π. He’d tried this crazy method throwing a lot of infinities and roots of infinities and imaginary numbers together. I believe I translated it into the language of modern mathematics fairly. And my failure is not that I found the formula actually described the number -½π.

Oh, I had an error in there, yes. And I’d found where it was. It was all the way back in the essay which first converted Wronski’s formula into something respectable. It was a small error, first appearing in the last formula of that essay and never corrected from there. This reinforces my suspicion that when normal people see formulas they mostly look at them to confirm there is a formula there. With luck they carry on and read the sentences around them.

My failure is I wanted to write a bit about boring mistakes. The kinds which you make all the time while doing mathematics work, but which you don’t worry about. Dropped signs. Constants which aren’t divided out, or which get multiplied in incorrectly. Stuff like this which you only detect because you know, deep down, that you should have gotten to an attractive simple formula and you haven’t. Mistakes which are tiresome to make, but never make you wonder if you’re in the wrong job.

The trouble is I can’t think of how to make an essay of that. We don’t tend to rate little mistakes like the wrong sign or the wrong multiple or a boring unnecessary added constant as important. This is because they’re not. The interesting stuff in a mathematical formula is usually the stuff representing variations. Change is interesting. The direction of the change? Eh, nice to know. A swapped plus or minus sign alters your understanding of the direction of the change, but that’s all. Multiplying or dividing by a constant wrongly changes your understanding of the size of the change. But that doesn’t alter what the change looks like. Just the scale of the change. Adding or subtracting the wrong constant alters what you think the change is varying from, but not what the shape of the change is. Once more, not a big deal.

But you also know that instinctively, or at least you get it from seeing how it’s worth one or two points on an exam to write -sin where you mean +sin. Or how if you ask the instructor in class about that 2 where a ½ should be, she’ll say, “Oh, yeah, you’re right” and do a hurried bit of erasing before going on.

Thus my failure: I don’t know what to say about boring mistakes that has any insight.

For the record here’s where I got things wrong. I was creating a function, named ‘f’ and using as a variable ‘x’, to represent Wronski’s formula. I’d gotten to this point:

$f(x) = -4 \imath x 2^{\frac{1}{2}\cdot \frac{1}{x}} \left\{ e^{\imath \frac{\pi}{4}\cdot\frac{1}{x}} - e^{- \imath \frac{\pi}{4}\cdot\frac{1}{x}} \right\}$

And then I observed how the stuff in curly braces there is “one of those magic tricks that mathematicians know because they see it all the time”. And I wanted to call in this formula, correctly:

$\sin\left(\phi\right) = \frac{e^{\imath \phi} - e^{-\imath \phi}}{2\imath }$

So here’s where I went wrong. I took the $4\imath$ way off in the front of that first formula and combined it with the stuff in braces to make 2 times a sine of some stuff. I apologize for this. I must have been writing stuff out faster than I was thinking about it. If I had thought, I would have gone through this intermediate step:

$f(x) = -4 \imath x 2^{\frac{1}{2}\cdot \frac{1}{x}} \left\{ e^{\imath \frac{\pi}{4}\cdot\frac{1}{x}} - e^{- \imath \frac{\pi}{4}\cdot\frac{1}{x}} \right\} \cdot \frac{2\imath}{2\imath}$

Because with that form in mind, it’s easy to take the stuff in curled braces and the $2\imath$ in the denominator. From that we get, correctly, $\sin\left(\frac{\pi}{4}\cdot\frac{1}{x}\right)$. And then the $-4\imath$ on the far left of that expression and the $2\imath$ on the right multiply together to produce the number 8.

So the function ought to have been, all along:

$f(x) = 8 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

Not very different, is it? Ah, but it makes a huge difference. Carry through with all the L’Hôpital’s Rule stuff described in previous essays. All the complicated formula work is the same. There’s a different number hanging off the front, waiting to multiply in. That’s all. And what you find, redoing all the work but using this corrected function, is that Wronski’s original mess —

$\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} - \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}$

— should indeed equal:

$2\pi$

All right, there’s an extra factor of 2 here. And I don’t think that is my mistake. Or if it is, other people come to the same mistake without my prompting.

Possibly the book I drew this from misquoted Wronski. It’s at least as good to have a formula for 2π as it is to have one for π. Or Wronski had a mistake in his original formula, and had a constant multiplied out front which he didn’t want. It happens to us all.

Fin.

## Wronski’s Formula For Pi: How Close We Came

Previously:

Józef Maria Hoëne-Wronski’s had an idea for a new, universal, culturally-independent definition of π. It was this formula that nobody went along with because they had looked at it:

$\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} - \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}$

I made some guesses about what he would want this to mean. And how we might put that in terms of modern, conventional mathematics. I describe those in the above links. In terms of limits of functions, I got this:

$\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

The trouble is that limit took more work than I wanted to do to evaluate. If you try evaluating that ‘f(x)’ at ∞, you get an expression that looks like zero times ∞. This begs for the use of L’Hôpital’s Rule, which tells you how to find the limit for something that looks like zero divided by zero, or like ∞ divided by ∞. Do a little rewriting — replacing that first ‘x’ with ‘$\frac{1}{1 / x}$ — and this ‘f(x)’ behaves like L’Hôpital’s Rule needs.

The trouble is, that’s a pain to evaluate. L’Hôpital’s Rule works on functions that look like one function divided by another function. It does this by calculating the derivative of the numerator function divided by the derivative of the denominator function. And I decided that was more work than I wanted to do.

Where trouble comes up is all those parts where $\frac{1}{x}$ turns up. The derivatives of functions with a lot of $\frac{1}{x}$ terms in them get more complicated than the original functions were. Is there a way to get rid of some or all of those?

And there is. Do a change of variables. Let me summon the variable ‘y’, whose value is exactly $\frac{1}{x}$. And then I’ll define a new function, ‘g(y)’, whose value is whatever ‘f’ would be at $\frac{1}{y}$. That is, and this is just a little bit of algebra:

$g(y) = -2 \cdot \frac{1}{y} \cdot 2^{\frac{1}{2} y } \cdot \sin\left(\frac{\pi}{4} y\right)$

The limit of ‘f(x)’ for ‘x’ at ∞ should be the same number as the limit of ‘g(y)’ for ‘y’ at … you’d really like it to be zero. If ‘x’ is incredibly huge, then $\frac{1}{x}$ has to be incredibly small. But we can’t just swap the limit of ‘x’ at ∞ for the limit of ‘y’ at 0. The limit of a function at a point reflects the value of the function at a neighborhood around that point. If the point’s 0, this includes positive and negative numbers. But looking for the limit at ∞ gets at only positive numbers. You see the difference?

… For this particular problem it doesn’t matter. But it might. Mathematicians handle this by taking a “one-sided limit”, or a “directional limit”. The normal limit at 0 of ‘g(y)’ is based on what ‘g(y)’ looks like in a neighborhood of 0, positive and negative numbers. In the one-sided limit, we just look at a neighborhood of 0 that’s all values greater than 0, or less than 0. In this case, I want the neighborhood that’s all values greater than 0. And we write that by adding a little + in superscript to the limit. For the other side, the neighborhood less than 0, we add a little – in superscript. So I want to evalute:

$\displaystyle \lim_{y \to 0^+} g(y) = \lim_{y \to 0^+} -2\cdot\frac{2^{\frac{1}{2}y} \cdot \sin\left(\frac{\pi}{4} y\right)}{y}$

Limits and L’Hôpital’s Rule and stuff work for one-sided limits the way they do for regular limits. So there’s that mercy. The first attempt at this limit, seeing what ‘g(y)’ is if ‘y’ happens to be 0, gives $-2 \cdot \frac{1 \cdot 0}{0}$. A zero divided by a zero is promising. That’s not defined, no, but it’s exactly the format that L’Hôpital’s Rule likes. The numerator is:

$-2 \cdot 2^{\frac{1}{2}y} \sin\left(\frac{\pi}{4} y\right)$

And the denominator is:

$y$

The first derivative of the denominator is blessedly easy: the derivative of y, with respect to y, is 1. The derivative of the numerator is a little harder. It demands the use of the Product Rule and the Chain Rule, just as last time. But these chains are easier.

The first derivative of the numerator is going to be:

$-2 \cdot 2^{\frac{1}{2}y} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} y\right) + -2 \cdot 2^{\frac{1}{2}y} \cdot \cos\left(\frac{\pi}{4} y\right) \cdot \frac{\pi}{4}$

Yeah, this is the simpler version of the thing I was trying to figure out last time. Because this is what’s left if I write the derivative of the numerator over the derivative of the denominator:

$\displaystyle \lim_{y \to 0^+} \frac{ -2 \cdot 2^{\frac{1}{2}y} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} y\right) + -2 \cdot 2^{\frac{1}{2}y} \cdot \cos\left(\frac{\pi}{4} y\right) \cdot \frac{\pi}{4} }{1}$

And now this is easy. Promise. There’s no expressions of ‘y’ divided by other expressions of ‘y’ or anything else tricky like that. There’s just a bunch of ordinary functions, all of them defined for when ‘y’ is zero. If this limit exists, it’s got to be equal to:

$\displaystyle -2 \cdot 2^{\frac{1}{2} 0} \cdot \log(2) \cdot \frac{1}{2} \cdot \sin\left(\frac{\pi}{4} \cdot 0\right) + -2 \cdot 2^{\frac{1}{2} 0 } \cdot \cos\left(\frac{\pi}{4} \cdot 0\right) \cdot \frac{\pi}{4}$

$\frac{\pi}{4} \cdot 0$ is 0. And the sine of 0 is 0. The cosine of 0 is 1. So all this gets to be a lot simpler, really fast.

$\displaystyle -2 \cdot 2^{0} \cdot \log(2) \cdot \frac{1}{2} \cdot 0 + -2 \cdot 2^{ 0 } \cdot 1 \cdot \frac{\pi}{4}$

And 20 is equal to 1. So the part to the left of the + sign there is all zero. What remains is:

$\displaystyle 0 + -2 \cdot \frac{\pi}{4}$

And so, finally, we have it. Wronski’s formula, as best I make it out, is a function whose value is …

$-\frac{\pi}{2}$

… So, what Wronski had been looking for, originally, was π. This is … oh, so very close to right. I mean, there’s π right there, it’s just multiplied by an unwanted $-\frac{1}{2}$. The question is, where’s the mistake? Was Wronski wrong to start with? Did I parse him wrongly? Is it possible that the book I copied Wronski’s formula from made a mistake?

Could be any of them. I’d particularly suspect I parsed him wrongly. I returned the library book I had got the original claim from, and I can’t find it again before this is set to publish. But I should check whether Wronski was thinking to find π, the ratio of the circumference to the diameter of a circle. Or might he have looked to find the ratio of the circumference to the radius of a circle? Either is an interesting number worth finding. We’ve settled on the circumference-over-diameter as valuable, likely for practical reasons. It’s much easier to measure the diameter than the radius of a thing. (Yes, I have read the Tau Manifesto. No, I am not impressed by it.) But if you know 2π, then you know π, or vice-versa.

The next question: yeah, but I turned up -½π. What am I talking about 2π for? And the answer there is, I’m not the first person to try working out Wronski’s stuff. You can try putting the expression, as best you parse it, into a tool like Mathematica and see what makes sense. Or you can read, for example, Quora commenters giving answers with way less exposition than I do. And I’m convinced: somewhere along the line I messed up. Not in an important way, but, essentially, doing something equivalent to divided by -2 when I should have multiplied by that.

I’ve spotted my mistake. I figure to come back around to explaining where it is and how I made it.

## Wronski’s Formula For Pi: Two Weird Tricks For Limits That Mathematicians Keep Using

Previously:

So now a bit more on Józef Maria Hoëne-Wronski’s attempted definition of π. I had got it rewritten to this form:

$\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

And I’d tried the first thing mathematicians do when trying to evaluate the limit of a function at a point. That is, take the value of that point and put it in whatever the formula is. If that formula evaluates to something meaningful, then that value is the limit. That attempt gave this:

$-2 \cdot \infty \cdot 1 \cdot 0$

Because the limit of ‘x’, for ‘x’ at ∞, is infinitely large. The limit of ‘$2^{\frac{1}{2}\cdot\frac{1}{x}}$‘ for ‘x’ at ∞ is 1. The limit of ‘$\sin(\frac{\pi}{4}\cdot\frac{1}{x})$ for ‘x’ at ∞ is 0. We can take limits that are 0, or limits that are some finite number, or limits that are infinitely large. But multiplying a zero times an infinity is dangerous. Could be anything.

Mathematicians have a tool. We know it as L’Hôpital’s Rule. It’s named for the French mathematician Guillaume de l’Hôpital, who discovered it in the works of his tutor, Johann Bernoulli. (They had a contract giving l’Hôpital publication rights. If Wikipedia’s right the preface of the book credited Bernoulli, although it doesn’t appear to be specifically for this. The full story is more complicated and ambiguous. The previous sentence may be said about most things.)

So here’s the first trick. Suppose you’re finding the limit of something that you can write as the quotient of one function divided by another. So, something that looks like this:

$\displaystyle \lim_{x \to a} \frac{h(x)}{g(x)}$

(Normally, this gets presented as ‘f(x)’ divided by ‘g(x)’. But I’m already using ‘f(x)’ for another function and I don’t want to muddle what that means.)

Suppose it turns out that at ‘a’, both ‘h(x)’ and ‘g(x)’ are zero, or both ‘h(x)’ and ‘g(x)’ are ∞. Zero divided by zero, or ∞ divided by ∞, looks like danger. It’s not necessarily so, though. If this limit exists, then we can find it by taking the first derivatives of ‘h’ and ‘g’, and evaluating:

$\displaystyle \lim_{x \to a} \frac{h'(x)}{g'(x)}$

That ‘ mark is a common shorthand for “the first derivative of this function, with respect to the only variable we have around here”.

This doesn’t look like it should help matters. Often it does, though. There’s an excellent chance that either ‘h'(x)’ or ‘g'(x)’ — or both — aren’t simultaneously zero, or ∞, at ‘a’. And once that’s so, we’ve got a meaningful limit. This doesn’t always work. Sometimes we have to use this l’Hôpital’s Rule trick a second time, or a third or so on. But it works so very often for the kinds of problems we like to do. Reaches the point that if it doesn’t work, we have to suspect we’re calculating the wrong thing.

But wait, you protest, reasonably. This is fine for problems where the limit looks like 0 divided by 0, or ∞ divided by ∞. What Wronski’s formula got me was 0 times 1 times ∞. And I won’t lie: I’m a little unsettled by having that 1 there. I feel like multiplying by 1 shouldn’t be a problem, but I have doubts.

That zero times ∞ thing, thought? That’s easy. Here’s the second trick. Let me put it this way: isn’t ‘x’ really the same thing as $\frac{1}{ 1 / x }$?

I expect your answer is to slam your hand down on the table and glare at my writing with contempt. So be it. I told you it was a trick.

And it’s a perfectly good one. And it’s perfectly legitimate, too. $\frac{1}{x}$ is a meaningful number if ‘x’ is any finite number other than zero. So is $\frac{1}{ 1 / x }$. Mathematicians accept a definition of limit that doesn’t really depend on the value of your expression at a point. So that $\frac{1}{x}$ wouldn’t be meaningful for ‘x’ at zero doesn’t mean we can’t evaluate its limit for ‘x’ at zero. And just because we might not be sure that $\frac{1}{x}$ would mean for infinitely large ‘x’ doesn’t mean we can’t evaluate its limit for ‘x’ at ∞.

I see you, person who figures you’ve caught me. The first thing I tried was putting in the value of ‘x’ at the ∞, all ready to declare that this was the limit of ‘f(x)’. I know my caveats, though. Plugging in the value you want the limit at into the function whose limit you’re evaluating is a shortcut. If you get something meaningful, then that’s the same answer you would get finding the limit properly. Which is done by looking at the neighborhood around but not at that point. So that’s why this reciprocal-of-the-reciprocal trick works.

So back to my function, which looks like this:

$\displaystyle f(x) = -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

Do I want to replace ‘x’ with $\frac{1}{1 / x}$, or do I want to replace $\sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$ with $\frac{1}{1 / \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)}$? I was going to say something about how many times in my life I’ve been glad to take the reciprocal of the sine of an expression of x. But just writing the symbols out like that makes the case better than being witty would.

So here is a new, L’Hôpital’s Rule-friendly, version of my version of Wronski’s formula:

$\displaystyle f(x) = -2 \frac{2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)}{\frac{1}{x}}$

I put that -2 out in front because it’s not really important. The limit of a constant number times some function is the same as that constant number times the limit of that function. We can put that off to the side, work on other stuff, and hope that we remember to bring it back in later. I manage to remember it about four-fifths of the time.

So these are the numerator and denominator functions I was calling ‘h(x)’ and ‘g(x)’ before:

$h(x) = 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

$g(x) = \frac{1}{x}$

The limit of both of these at ∞ is 0, just as we might hope. So we take the first derivatives. That for ‘g(x)’ is easy. Anyone who’s reached week three in Intro Calculus can do it. This may only be because she’s gotten bored and leafed through the formulas on the inside front cover of the textbook. But she can do it. It’s:

$g'(x) = -\frac{1}{x^2}$

The derivative for ‘h(x)’ is a little more involved. ‘h(x)’ we can write as the product of two expressions, that $2^{\frac{1}{2}\cdot \frac{1}{x}}$ and that $\sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$. And each of those expressions contains within themselves another expression, that $\frac{1}{x}$. So this is going to require the Product Rule, of two expressions that each require the Chain Rule.

This is as far as I got with that before slamming my hand down on the table and glaring at the problem with disgust:

$h'(x) = 2^{\frac{1}{2}\frac{1}{x}} \cdot \log(2) \cdot \frac{1}{2} \cdot (-1) \cdot \frac{1}{x^2} + 2^{\frac{1}{2}\frac{1}{x}} \cdot \cos( arg ) bleah$

Yeah I’m not finishing that. Too much work. I’m going to reluctantly try thinking instead.

(If you want to do that work — actually, it isn’t much more past there, and if you followed that first half you’re going to be fine. And you’ll see an echo of it in what I do next time.)

## Wronski’s Formula For Pi: A First Limit

Previously:

When I last looked at Józef Maria Hoëne-Wronski’s attempted definition of π I had gotten it to this. Take the function:

$f(x) = -2 x 2^{\frac{1}{2}\cdot \frac{1}{x}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{x}\right)$

And find its limit when ‘x’ is ∞. Formally, you want to do this by proving there’s some number, let’s say ‘L’. And ‘L’ has the property that you can pick any margin-of-error number ε that’s bigger than zero. And whatever that ε is, there’s some number ‘N’ so that whenever ‘x’ is bigger than ‘N’, ‘f(x)’ is larger than ‘L – ε’ and also smaller than ‘L + ε’. This can be a lot of mucking about with expressions to prove.

Fortunately we have shortcuts. There’s work we can do that gets us ‘L’, and we can rely on other proofs that show that this must be the limit of ‘f(x)’ at some value ‘a’. I use ‘a’ because that doesn’t commit me to talking about ∞ or any other particular value. The first approach is to just evaluate ‘f(a)’. If you get something meaningful, great! We’re done. That’s the limit of ‘f(x)’ at ‘a’. This approach is called “substitution” — you’re substituting ‘a’ for ‘x’ in the expression of ‘f(x)’ — and it’s great. Except that if your problem’s interesting then substitution won’t work. Still, maybe Wronski’s formula turns out to be lucky. Fit in ∞ where ‘x’ appears and we get:

$f(\infty) = -2 \infty 2^{\frac{1}{2}\cdot \frac{1}{\infty}} \sin\left(\frac{\pi}{4}\cdot \frac{1}{\infty}\right)$

So … all right. Not quite there yet. But we can get there. For example, $\frac{1}{\infty}$ has to be — well. It’s what you would expect if you were a kid and not worried about rigor: 0. We can make it rigorous if you like. (It goes like this: Pick any ε larger than 0. Then whenever ‘x’ is larger than $\frac{1}{\epsilon}$ then $\frac{1}{x}$ is less than ε. So the limit of $\frac{1}{x}$ at ∞ has to be 0.) So let’s run with this: replace all those $\frac{1}{\infty}$ expressions with 0. Then we’ve got:

$f(\infty) = -2 \infty 2^{0} \sin\left(0\right)$

The sine of 0 is 0. 20 is 1. So substitution tells us limit is -2 times ∞ times 1 times 0. That there’s an ∞ in there isn’t a problem. A limit can be infinitely large. Think of the limit of ‘x2‘ at ∞. An infinitely large thing times an infinitely large thing is fine. The limit of ‘x ex‘ at ∞ is infinitely large. A zero times a zero is fine; that’s zero again. But having an ∞ times a 0? That’s trouble. ∞ times something should be huge; anything times zero should be 0; which term wins?

So we have to fall back on alternate plans. Fortunately there’s a tool we have for limits when we’d otherwise have to face an infinitely large thing times a zero.

I hope to write about this next time. I apologize for not getting through it today but time wouldn’t let me.

## As I Try To Figure Out What Wronski Thought ‘Pi’ Was

A couple weeks ago I shared a fascinating formula for π. I got it from Carl B Boyer’s The History of Calculus and its Conceptual Development. He got it from Józef Maria Hoëne-Wronski, early 19th-century Polish mathematician. His idea was that an absolute, culturally-independent definition of π would come not from thinking about circles and diameters but rather this formula:

$\pi = \frac{4\infty}{\sqrt{-1}}\left\{ \left(1 + \sqrt{-1}\right)^{\frac{1}{\infty}} - \left(1 - \sqrt{-1}\right)^{\frac{1}{\infty}} \right\}$

Now, this formula is beautiful, at least to my eyes. It’s also gibberish. At least it’s ungrammatical. Mathematicians don’t like to write stuff like “four times infinity”, at least not as more than a rough draft on the way to a real thought. What does it mean to multiply four by infinity? Is arithmetic even a thing that can be done on infinitely large quantities? Among Wronski’s problems is that they didn’t have a clear answer to this. We’re a little more advanced in our mathematics now. We’ve had a century and a half of rather sound treatment of infinitely large and infinitely small things. Can we save Wronski’s work?

Start with the easiest thing. I’m offended by those $\sqrt{-1}$ bits. Well, no, I’m more unsettled by them. I would rather have $\imath$ in there. The difference? … More taste than anything sound. I prefer, if I can get away with it, using the square root symbol to mean the positive square root of the thing inside. There is no positive square root of -1, so, pfaugh, away with it. Mere style? All right, well, how do you know whether those $\sqrt{-1}$ terms are meant to be $\imath$ or its additive inverse, $-\imath$? How do you know they’re all meant to be the same one? See? … As with all style preferences, it’s impossible to be perfectly consistent. I’m sure there are times I accept a big square root symbol over a negative or a complex-valued quantity. But I’m not forced to have it here so I’d rather not. First step:

$\pi = \frac{4\infty}{\imath}\left\{ \left(1 + \imath\right)^{\frac{1}{\infty}} - \left(1 - \imath\right)^{\frac{1}{\infty}} \right\}$

Also dividing by $\imath$ is the same as multiplying by $-\imath$ so the second easy step gives me:

$\pi = -4 \imath \infty \left\{ \left(1 + \imath\right)^{\frac{1}{\infty}} - \left(1 - \imath\right)^{\frac{1}{\infty}} \right\}$

Now the hard part. All those infinities. I don’t like multiplying by infinity. I don’t like dividing by infinity. I really, really don’t like raising a quantity to the one-over-infinity power. Most mathematicians don’t. We have a tool for dealing with this sort of thing. It’s called a “limit”.

Mathematicians developed the idea of limits over … well, since they started doing mathematics. In the 19th century limits got sound enough that we still trust the idea. Here’s the rough way it works. Suppose we have a function which I’m going to name ‘f’ because I have better things to do than give functions good names. Its domain is the real numbers. Its range is the real numbers. (We can define functions for other domains and ranges, too. Those definitions look like what they do here.)

I’m going to use ‘x’ for the independent variable. It’s any number in the domain. I’m going to use ‘a’ for some point. We want to know the limit of the function “at a”. ‘a’ might be in the domain. But — and this is genius — it doesn’t have to be. We can talk sensibly about the limit of a function at some point where the function doesn’t exist. We can say “the limit of f at a is the number L”. I hadn’t introduced ‘L’ into evidence before, but … it’s a number. It has some specific set value. Can’t say which one without knowing what ‘f’ is and what its domain is and what ‘a’ is. But I know this about it.

Pick any error margin that you like. Call it ε because mathematicians do. However small this (positive) number is, there’s at least one neighborhood in the domain of ‘f’ that surrounds ‘a’. Check every point in that neighborhood other than ‘a’. The value of ‘f’ at all those points in that neighborhood other than ‘a’ will be larger than L – ε and smaller than L + ε.

Yeah, pause a bit there. It’s a tricky definition. It’s a nice common place to crash hard in freshman calculus. Also again in Intro to Real Analysis. It’s not just you. Perhaps it’ll help to think of it as a kind of mutual challenge game. Try this.

1. You draw whatever error bar, as big or as little as you like, around ‘L’.
2. But I always respond by drawing some strip around ‘a’.
3. You then pick absolutely any ‘x’ inside my strip, other than ‘a’.
4. Is f(x) always within the error bar you drew?

Suppose f(x) is. Suppose that you can pick any error bar however tiny, and I can answer with a strip however tiny, and every single ‘x’ inside my strip has an f(x) within your error bar … then, L is the limit of f at a.

Again, yes, tricky. But mathematicians haven’t found a better definition that doesn’t break something mathematicians need.

To write “the limit of f at a is L” we use the notation:

$\displaystyle \lim_{x \to a} f(x) = L$

The ‘lim’ part probably makes perfect sense. And you can see where ‘f’ and ‘a’ have to enter into it. ‘x’ here is a “dummy variable”. It’s the falsework of the mathematical expression. We need some name for the independent variable. It’s clumsy to do without. But it doesn’t matter what the name is. It’ll never appear in the answer. If it does then the work went wrong somewhere.

What I want to do, then, is turn all those appearances of ‘∞’ in Wronski’s expression into limits of something at infinity. And having just said what a limit is I have to do a patch job. In that talk about the limit at ‘a’ I talked about a neighborhood containing ‘a’. What’s it mean to have a neighborhood “containing ∞”?

The answer is exactly what you’d think if you got this question and were eight years old. The “neighborhood of infinity” is “all the big enough numbers”. To make it rigorous, it’s “all the numbers bigger than some finite number that let’s just call N”. So you give me an error bar around ‘L’. I’ll give you back some number ‘N’. Every ‘x’ that’s bigger than ‘N’ has f(x) inside your error bars. And note that I don’t have to say what ‘f(∞)’ is or even commit to the idea that such a thing can be meaningful. I only ever have to think directly about values of ‘f(x)’ where ‘x’ is some real number.

So! First, let me rewrite Wronski’s formula as a function, defined on the real numbers. Then I can replace each ∞ with the limit of something at infinity and … oh, wait a minute. There’s three ∞ symbols there. Do I need three limits?

Ugh. Yeah. Probably. This can be all right. We can do multiple limits. This can be well-defined. It can also be a right pain. The challenge-and-response game needs a little modifying to work. You still draw error bars. But I have to draw multiple strips. One for each of the variables. And every combination of values inside all those strips has give an ‘f’ that’s inside your error bars. There’s room for great mischief. You can arrange combinations of variables that look likely to break ‘f’ outside the error bars.

So. Three independent variables, all taking a limit at ∞? That’s not guaranteed to be trouble, but I’d expect trouble. At least I’d expect something to keep the limit from existing. That is, we could find there’s no number ‘L’ so that this drawing-neighborhoods thing works for all three variables at once.

Let’s try. One of the ∞ will be a limit of a variable named ‘x’. One of them a variable named ‘y’. One of them a variable named ‘z’. Then:

$f(x, y, z) = -4 \imath x \left\{ \left(1 + \imath\right)^{\frac{1}{y}} - \left(1 - \imath\right)^{\frac{1}{z}} \right\}$

Without doing the work, my hunch is: this is utter madness. I expect it’s probably possible to make this function take on many wildly different values by the judicious choice of ‘x’, ‘y’, and ‘z’. Particularly ‘y’ and ‘z’. You maybe see it already. If you don’t, you maybe see it now that I’ve said you maybe see it. If you don’t, I’ll get there, but not in this essay. But let’s suppose that it’s possible to make f(x, y, z) take on wildly different values like I’m getting at. This implies that there’s not any limit ‘L’, and therefore Wronski’s work is just wrong.

Thing is, Wronski wouldn’t have thought that. Deep down, I am certain, he thought the three appearances of ∞ were the same “value”. And that to translate him fairly we’d use the same name for all three appearances. So I am going to do that. I shall use ‘x’ as my variable name, and replace all three appearances of ∞ with the same variable and a common limit. So this gives me the single function:

$f(x) = -4 \imath x \left\{ \left(1 + \imath\right)^{\frac{1}{x}} - \left(1 - \imath\right)^{\frac{1}{x}} \right\}$

And then I need to take the limit of this at ∞. If Wronski is right, and if I’ve translated him fairly, it’s going to be π. Or something easy to get π from.

I hope to get there next week.

## L’Hopital’s Rule Without End: Is That A Thing?

I was helping a friend learn L’Hôpital’s Rule. This is a Freshman Calculus thing. (A different one from last week, it happens. Folks are going back to school, I suppose.) The friend asked me a point I thought shouldn’t come up. I’m certain it won’t come up in the exam my friend was worried about, but I couldn’t swear it wouldn’t happen at all. So this is mostly a note to myself to think it over and figure out whether the trouble could come up. And also so this won’t be my most accessible post; I’m sorry for that, for folks who aren’t calculus-familiar.

L’Hôpital’s Rule is a way of evaluating the limit of one function divided by another, of f(x) divided by g(x). If the limit of $\frac{f(x)}{g(x)}$ has either the form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$ then you’re not stuck. You can take the first derivative of the numerator and the denominator separately. The limit of $\frac{f'(x)}{g'(x)}$ if it exists will be the same value.

But it’s possible to have to do this several times over. I used the example of finding the limit, as x grows infinitely large, where f(x) = x2 and g(x) = ex. $\frac{x^2}{e^x}$ goes to $\frac{\infty}{\infty}$ as x grows infinitely large. The first derivatives, $\frac{2x}{e^x}$, also go to $\frac{\infty}{\infty}$. You have to repeat the process again, taking the first derivatives of numerator and denominator again. $\frac{2}{e^x}$ finally goes to 0 as x gets infinitely large. You might have to do this a bunch of times. If f(x) were x7 and g(x) again ex you’d properly need to do this seven times over. With experience you figure out you can skip some steps. Of course students don’t have the experience to know they can skip ahead to the punch line there, but that’s what the practice in homework is for.

Anyway, my friend asked whether it’s possible to get a pattern that always ends up with $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and never breaks out of this. And that’s what’s got me stuck. I can think of a few patterns that would. Start out, for example, with f(x) = e3x and g(x) = e2x. Properly speaking, that would never end. You’d get an infinity-over-infinity pattern every derivative you took. Similarly, if you started with $f(x) = \frac{1}{x}$ and $g(x) = e^{-x}$ you’d never come to an end. As x got infinitely large both f(x) and g(x) would go to zero and all their derivatives would be zero over and over and over and over again.

But those are special cases. Anyone looking at what they were doing instead of just calculating would look at, say, $\frac{e^{3x}}{e^{2x}}$ and realize that’s the same as $e^x$ which falls out of the L’Hôpital’s Rule formulas. Or $\frac{\frac{1}{x}}{e^{-x}}$ would be the same as $\frac{e^x}{x}$ which is an infinity-over-infinity form. But it takes only one derivative to break out of the infinity-over-infinity pattern.

So I can construct examples that never break out of a zero-over-zero or an infinity-over-infinity pattern if you calculate without thinking. And calculating without thinking is a common problem students have. Arguably it’s the biggest problem mathematics students have. But what I wonder is, are there ratios that end up in an endless zero-over-zero or infinity-over-infinity pattern even if you do think it out?

And thus this note; I’d like to nag myself into thinking about that.

## Also, Some Mathematics Answers, Whatever Those Are

Solved Problems In Mathematics appears to be a new mathematics blog. It’s a few posts old, but I’m charmed by it. Each post takes some problem and shows how to solve it. It’s been heavy on calculus so far. Many limits, a bunch of evaluate-the-integral problems, and so on. But it may be a good bit of reading if you want to brush up on stuff you once knew, like how to calculate the length of peculiar curves. Or if you have a calculus problem you need to solve and haven’t got any idea what to do. A similar problem may give you a good hint.

Less generally, John Quintanilla’s Mean Green Math has looked at one specific calculus problem, a limit, in several different ways. He’s grouped the essays about that together. Most mathematics problems, once you’re away from the cutting edge of human knowledge, have many different possible solutions. Here’s an example of such.

And, what the heck. Math With Bad Drawings hardly needs my publicity. But a recent post discusses the challenge of saying just what mathematics is. The comments get into some decent discussion about what mathematics is. And I’m not sure what mathematics is. There are many facets which seem barely to have anything to do with one another; but they do all feel related in some way. It’s hard to say just what unites all this.

## Error

This is one of my A to Z words that everyone knows. An error is some mistake, evidence of our human failings, to be minimized at all costs. That’s … well, it’s an attitude that doesn’t let you use error as a tool.

An error is the difference between what we would like to know and what we do know. Usually, what we would like to know is something hard to work out. Sometimes it requires complicated work. Sometimes it requires an infinite amount of work to get exactly right. Who has the time for that?

This is how we use errors. We look for methods that approximate the thing we want, and that estimate how much of an error that method makes. Usually, the method involves doing some basic step some large number of times. And usually, if we did the step more times, the estimate of the error we make will be smaller. My essay “Calculating Pi Less Terribly” shows an example of this. If we add together more terms from that Leibniz formula we get a running total that’s closer to the actual value of π.

## Quick Little Calculus Puzzle

fluffy, one of my friends and regular readers, got to discussing with me a couple of limit problems, particularly, ones that seemed to be solved through L’Hopital’s Rule and then ran across some that don’t call for that tool of Freshman Calculus which you maybe remember. It’s the thing about limits of zero divided by zero, or infinity divided by infinity. (It can also be applied to a couple of other “indeterminate forms”; I remember when I took this level calculus the teacher explaining there were seven such forms. Without looking them up, I think they’re $\frac00, \frac{\infty}{\infty}, 0^0, \infty^{0}, 0^{\infty}, 1^{\infty}, \mbox{ and } \infty - \infty$ but I would not recommend trusting my memory in favor of actually studying for your test.)

Anyway, fluffy put forth two cute little puzzles that I had immediate responses for, and then started getting plagued by doubts about, so I thought I’d put them out here for people who want the recreation. They’re both about taking the limit at zero of fractions, specifically:

$\lim_{x \rightarrow 0} \frac{e^x}{x^e}$

$\lim_{x \rightarrow 0} \frac{x^e}{e^x}$

where e here is the base of the natural logarithm, that is, that number just a little high of 2.71828 that mathematicians find so interesting even though it isn’t pi.

The limit is, if you want to be exact, a subtly and carefully defined idea that took centuries of really bright work to explain. But the first really good feeling that I really got for it is to imagine a function evaluated at the points near but not exactly at the target point — in the limits here, where x equals zero — and to see, if you keep evaluating x very near zero, are the values of your expression very near something? If it does, that thing the expression gets near is probably the limit at that point.

So, yes, you can plug in values of x like 0.1 and 0.01 and 0.0001 and so on into $\frac{e^x}{x^e}$ and $\frac{x^e}{e^x}$ and get a feeling for what the limit probably is. Saying what it definitely is takes a little more work.